For $X_1,X_2,X_3 \sim$i.i.d exponential ($\lambda$), I am trying to show independence between $X_{(3)}$ and $X_{(2)}-X_{(3)}$ where $X_{(3)}$ is the third largest observation, i.e. the minimum in this case.
My attempt: \begin{align} P(X_{(2)}-X_{(3)} > x,X_{(3)} >y ) &= P(X_{(2)}-X_{(3)} > x \mid X_{(3)} >y )P(X_{(3)} >y) \\ & = P(X_{(2)}-X_{(3)} > x-y)P(X_{(3)} >y) \end{align} By the memorylessness of the exponential. I then go on to derive the the PDF of $X_{(2)}-X_{(3)}$ and show that the above expression is equivalent to $P(X_{(2)}-X_{(3)} > x)P(X_{(3)} >y)$ implying independence. However, this is quite tedious as it requires a lot of computation to figure out the pdf. I think there is a more elegant way to do it using the properties of the exponential but not quite sure how to go about it
Let $\lambda=1$ for simplicity and $X_{(1)}=\min\{X_1,X_2,X_3\}$. Then for $x\le y$
$$f_{X_{(1)},X_{(2)}}(x,y)=3!(1-F_{X_1}(y))f_{X_1}(x)f_{X_1}(y)$$
$X_{(2)}$ given $X_{(1)}$ is the sum of $X_{(1)}$ and an independent exponential r.v. with parameter 2 because
$$f_{X_{(2)}|X_{(1)}}(y|x)=\frac{3!(1-F_{X_1}(y))f_{X_1}(x)f_{X_1}(y)}{3(1-F_{X_1}(x))^2f_{X_1}(x)}=2e^{-2(y-x)}$$
so that
$$f_{X_{(2)}-X_{(1)}|X_{(1)}}(y|x)=2e^{-2y}=f_{X_{(2)}-X_{(1)}}(y)$$